$\lim _{n \rightarrow \infty} \frac{(2n(2n-1) \dots (n+1))^{1/n}}{n} = $

  • A
    $\int_0^1 \ln x \, dx$
  • B
    $\int_0^1 x \ln x \, dx$
  • C
    $\int_0^1 (x+1) \ln (x+1) \, dx$
  • D
    $\int_0^1 \ln (1+x) \, dx$

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$\lim _{n \rightarrow \infty}\left[\frac{\sqrt{n^2-1^2}}{n^2}+\frac{\sqrt{n^2-2^2}}{n^2}+\frac{\sqrt{n^2-3^2}}{n^2}+\ldots+\frac{\sqrt{n^2-n^2}}{n^2}\right]=$

$a \in R, |a| > 1$ के लिए,मान लीजिए $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. तो $a$ का/के संभावित मान है/हैं:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$

$\lim _{n \rightarrow \infty} \frac{3}{n} \left\{ 4 + \left( 2 + \frac{1}{n} \right)^2 + \left( 2 + \frac{2}{n} \right)^2 + \dots + \left( 3 - \frac{1}{n} \right)^2 \right\}$ का मान ज्ञात कीजिए।

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